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## rainbow22 3 years ago Find the limit to 0 of.... [(e^2x)-1]/x Is it possible to do this w/out a calculator?

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1. zzr0ck3r

do you know l`hopital rule yet?

2. rainbow22

no... :(

3. rainbow22

You can't have a zero in the denominater. The answer is 2

4. zzr0ck3r

dont just give answer, help them find answer....I dont know how to solve this wthout lhopitals rule

5. LolWolf

@zzr0ck3r yeah, I'm trying to find a direct way to do this, without l'Hopital's

6. rainbow22

idk, I'm just practicing for a test and this question was online...

7. zzr0ck3r

you nee to know how to tke a derivative to do this, you will lear that soon

8. rainbow22

I know how to take a derivative, but how will that help me solve it?

9. zzr0ck3r

take the derivative of the num and the the denom. then run the limit

10. zzr0ck3r

you will have (2e^2x)/1 now run the limi

11. zzr0ck3r

limit*

12. LolWolf

Oh, I've got one: Well we know: $e^x-1\sim x\\ x\to 0$ So: $\lim_{x\to 0}\frac{e^{2x}-1}{x}=\lim_{u\to 0}\frac{e^{u}-1}{\frac{1}{2}u}$Where $$u=2x$$ Due to that: $\lim_{x\to 0}\frac{u}{\frac{1}{2}u}=\frac{1}{\frac{1}{2}}=2$

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